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Analog Electronics circuits 10ES32

Department of EEE, SJBIT

ANALOG ELECTRONIC CIRCUITS

code:10ES32 IA marks:25 exam marks:100

UNIT 1: Diode Circuits: Diode Resistance, Diode equivalent circuits, Transition and diffusion capacitance,

Reverse recovery time, Load line analysis, Rectifiers, Clippers and clampers. 6 Hours

UNIT 2:

Transistor Biasing: Operating point, Fixed bias circuits, Emitter stabilized biased circuits, Voltage

dividerbiased, DC bias with voltage feedback, Miscellaneous bias configurations, Design operations,

Transistorswitching networks, PNP transistors, Bias stabilization. 6 Hours

UNIT 3:

Transistor at Low Frequencies: BJT transistor modeling, CE Fixed bias configuration, Voltage dividerbias, Emitter follower, CB configuration, Collector feedback configuration, Analysis of circuits

re model;analysis of CE configuration using h- parameter model; Relationship between h-parameter

model of CE,CCand CE configuration. 7 Hours

UNIT 4:

Transistor Frequency Response: General frequency considerations, low frequency response, Miller

effectcapacitance, High frequency response, multistage frequency effects. 7Hours

UNIT 5:

(a) General Amplifiers: Cascade connections, Cascode connections, Darlington connections.

3 Hours

(b) Feedback Amplifier: Feedback concept, Feedback connections type, Practical feedback

circuits.Design procedures for the feedback amplifiers.

4 Hours

UNIT 6:

Power Amplifiers: Definitions and amplifier types, series fed class A amplifier, Transformer

coupledClass A amplifiers, Class B amplifier operations, Class B amplifier circuits, Amplifier distortions.Designing ofPower amplifiers. 7 Hours

UNIT 7: Oscillators: Oscillator operation, Phase shift Oscillator, Wienbridge Oscillator, Tuned Oscillator

circuits,Crystal Oscillator. (BJT Version Only)Simple design methods of Oscillators. 6 Hours

UNIT 8: FET Amplifiers: FET small signal model, Biasing of FET, Common drain common gate

configurations,MOSFETs, FET amplifier networks. 6 Hours



Sl.no Contents Page

number

1 Unit-1: pn junction diode 3 to 23

Diode Resistance

Diode equivalent circuits

Transition and diffusion capacitance

Reverse recovery time

Load line analysis.

Rectifiers

Clippers

Clampers

2 Unit-2: DC Biasing of BJT 24 to 40

Operating point

Fixed bias circuits

Emitter stabilized biased circuits

Voltage divider biased

DC bias with voltage feedback

Miscellaneous bias configuration

Transistor switching networks

PNP transistors

3 Unit-3: Uni-junction transistor 41 to 48

BJT transistor modeling

CE Fixed bias configuration

Voltage dividerbias

Emitter follower



CB configuration

Collector feedback configuration

Analysis of circuits re model

analysis of CE configuration using h- parameter model

Relationship between h-parameter model of CE,CCand

CE configuration

4 Unit-4: Frequency response of Amplifiers 49 to 60

General frequency considerations

low frequency response

High frequency response

multistage frequency effects

5 Unit-5: General Amplifiers 61 to 78

Cascade connections

Cascode connections

Darlington connections.

Design procedures for the feedback amplifiers

6 Unit-6: Power amplifier 79 to 99

Definitions and amplifier types

series fed class A amplifier

Class A amplifiers,

Class B amplifier

Class C amplifier

Push Pull amplifier

7 Unit-7: Oscillator 100 to 108

Oscillator operation



Phase shift Oscillator

Wienbridge Oscillator

Crystal Oscillator

8 Unit-8: Field Effect Transistor 109 to 129

FET small signal model

Biasing of FET

FET amplifier networks.



Unit-1

p-n Junction Diode

Diode:

A pure silicon crystal or germanium crystal is known as an intrinsic semiconductor. There are

not enough free electrons and holes in an intrinsic semi-conductor to produce a usable

current. The electrical action of these can be modified by doping means adding impurity

atoms to a crystal to increase either the number of free holes or no of free electrons.

When a crystal has been doped, it is called a extrinsic semi-conductor. They are of two types

• n-type semiconductor having free electrons as majority carriers

• p-type semiconductor having free holes as majority carriers

By themselves, these doped materials are of little use. However, if a junction is made by

joining p-type semiconductor to n-type semiconductor a useful device is produced known as

diode. It will allow current to flow through it only in one direction. The unidirectional

properties of a diode allow current flow when forward biased and disallow current flow

when reversed biased. This is called rectification process and therefore it is also called

rectifier.

How is it possible that by properly joining two semiconductors each of which, by itself, will

freely conduct the current in any direct refuses to allow conduction in one direction.

Consider first the condition of p-type and n-type germanium just prior to joining fig. 1. The

majority and minority carriers are in constant motion.

The minority carriers are thermally produced and they exist only for short time after which

they recombine and neutralize each other. In the mean time, other minority carriers have

been produced and this process goes on and on.

The number of these electron hole pair that exist at any one time depends upon the

temperature. The number of majority carriers is however, fixed depending on the number of

impurity atoms available. While the electrons and holes are in motion but the atoms are fixed

in place and do not move.

http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-ROORKEE/BASIC-ELECTRONICS/lecturers/lecture_1/lecture1_page1.htm



Holes from the p-side diffuse into n-side where they recombine with free electrons.

Free electrons from n-side diffuse into p-side where they recombine with free holes.

The diffusion of electrons and holes is due to the fact that large no of electrons are

concentrated in one area and large no of holes are concentrated in another area.

When these electrons and holes begin to diffuse across the junction then they collide

each other and negative charge in the electrons cancels the positive charge of the hole and

both will lose their charges.

The diffusion of holes and electrons is an electric current referred to as a recombination

current. The recombination process decay exponentially with both time and distance from

the junction. Thus most of the recombination occurs just after the junction is made and very

near to junction.

A measure of the rate of recombination is the lifetime defined as the time required for

the density of carriers to decrease to 37% to the original concentration

The impurity atoms are fixed in their individual places. The atoms itself is a part of the

crystal and so cannot move. When the electrons and hole meet, their individual charge is

cancelled and this leaves the originating impurity atoms with a net charge, the atom that

produced the electron now lack an electronic and so becomes charged positively, whereas

the atoms that produced the hole now lacks a positive charge and becomes negative.

The electrically charged atoms are called ions since they are no longer neutral. These

ions produce an electric field as shown in fig. 3. After several collisions occur, the electric

field is great enough to repel rest of the majority carriers away of the junction. For example,

an electron trying to diffuse from n to p side is repelled by the negative charge of the p-side.

Thus diffusion process does not continue indefinitely but continues as long as the field is

developed.

This region is produced immediately surrounding the junction that has no majority

carriers. The majority carriers have been repelled away from the junction and junction is

depleted from carriers. The junction is known as the barrier region or depletion region. The

electric field represents a potential difference across the junction also called space charge

potential or barrier potential . This potential is 0.7v for Si at 25o celcious and 0.3v for Ge.




The physical width of the depletion region depends on the doping level. If very heavy

doping is used, the depletion region is physically thin because diffusion charge need not

travel far across the junction before recombination takes place (short life time). If doping is

light, then depletion is more wide (long life time).

The symbol of diode is shown in fig. 4. The terminal connected to p-layer is called anode (A)

and the terminal connected to n-layer is called cathode (K)

Fig.4

Reverse Bias:

If positive terminal of dc source is connected to cathode and negative terminal is connected to

anode, the diode is called reverse biased as shown in fig. 5.

Space charge capacitance CT of diode:

Reverse bias causes majority carriers to move away from the junction, thereby creating

more ions. Hence the thickness of depletion region increases. This region behaves as the

dielectric material used for making capacitors. The p-type and n-type conducting on each

side of dielectric act as the plate. The incremental capacitance CT is defined by

Since

Therefore, (E-1)





where, dQ is the increase in charge caused by a change dV in voltage. CT is not constant, it

depends upon applied voltage, there fore it is defined as dQ / dV.

When p-n junction is forward biased, then also a capacitance is defined called diffusion

capacitance CD (rate of change of injected charge with voltage) to take into account the time

delay in moving the charges across the junction by the diffusion process. It is considered as

a fictitious element that allow us to predict time delay.

If the amount of charge to be moved across the junction is increased, the time delay is greater,

it follows that diffusion capacitance varies directly with the magnitude of forward current.

(E-2)

Relationship between Diode Current and Diode Voltage

An exponential relationship exists between the carrier density and applied potential of diode

junction as given in equation E-3. This exponential relationship of the current iD and the

voltage vD holds over a range of at least seven orders of magnitudes of current - that is a

factor of 107.

(E-3)

Where,

iD= Current through the diode (dependent variable in this expression)

vD= Potential difference across the diode terminals (independent variable in this expression)

IO= Reverse saturation current (of the order of 10-15

A for small signal diodes, but IO is a

strong function of temperature)

q = Electron charge: 1.60 x 10-19

joules/volt

k = Boltzmann's constant: 1.38 x l0-23

joules /° K

T = Absolute temperature in degrees Kelvin (°K = 273 + temperature in °C)

n = Empirical scaling constant between 0.5 and 2, sometimes referred to as the Exponential

Ideality Factor



The empirical constant, n, is a number that can vary according to the voltage and current levels.

It depends on electron drift, diffusion, and carrier recombination in the depletion region.

Among the quantities affecting the value of n are the diode manufacture, levels of doping

and purity of materials. If n=1, the value of k T/ q is 26 mV at 25°C. When n=2, k T/ q

becomes 52 mV.

For germanium diodes, n is usually considered to be close to 1. For silicon diodes, n is in the

range of 1.3 to 1.6. n is assumed 1 for all junctions all throughout unless otherwise noted.

Equation (E-3) can be simplified by defining VT =k T/q, yielding

(E-4)

At room temperature (25°C) with forward-bias voltage only the first term in the parentheses is

dominant and the current is approximately given by

(E-5)

Fig.1 - Diode Voltage relationship

The slope of the curves of fig.1 changes as the current and voltage change since the l-V

characteristic follows the exponential relationship of relationship of equation (E-4).

Differentiate the equation (E-4) to find the slope at any arbitrary value of vDor iD,

(E-6)




This slope is the equivalent conductance of the diode at the specified values of vD or iD.

We can approximate the slope as a linear function of the diode current. To eliminate the

exponential function, we substitute equation (E-4) into the exponential of equation (E-7) to

obtain

(E-7)

A realistic assumption is that IO<< iD equation (E-7) then yields,

(E-8)

The approximation applies if the diode is forward biased. The dynamic resistance is the

reciprocal of this expression.

(E-9)

Although rd is a function of id, we can approximate it as a constant if the variation of iD is

small. This corresponds to approximating the exponential function as a straight line within a

specific operating range.

Normally, the term Rf to denote diode forward resistance. Rf is composed of rd and the contact

resistance. The contact resistance is a relatively small resistance composed of the resistance

of the actual connection to the diode and the resistance of the semiconductor prior to the

junction. The reverse-bias resistance is extremely large and is often approximated as

infinity.

The slope of this line is equal to 1/ RL. The other equation in terms of these two variables

VD & Id, is given by the static characteristic. The point of intersection of straight line

and diode characteristic gives the operating point as shown in fig. 4.




ohm.

The resulting input voltage is the sum of dc voltage and sinusoidal ac voltage. Therefore,

as the diode voltage varies, diode current also varies, sinusoidally. The intersection of

load line and diode characteristic for different input voltages gives the output voltage

as shown in fig. 6.




Fig. 6

In certain applications only ac equivalent circuit is required. Since only ac response of the

circuit is considered DC Source is not shown in the equivalent circuit of fig. 7. The

resistance rf represents the dynamic resistance or ac resistance of the diode. It is

obtained by taking the ratio of Δ VD/ Δ ID at operating point.

Dynamic Resistance Δ rD = Δ VD / Δ ID

Let us consider a circuit shown in fig. 5 having dc voltage and sinusoidal ac voltage. Say





V = 1V, RL=10

Fig. 7

Applications of Diode

Half wave Rectifier:

The single ? phase half wave rectifier is shown in fig. 8.

Fig. 8 Fig. 9

In positive half cycle, D is forward biased and conducts. Thus the output voltage is same as the

input voltage. In the negative half cycle, D is reverse biased, and therefore output voltage is

zero. The output voltage waveform is shown in fig. 9.

The average output voltage of the rectifier is given by





The average output current is given by

When the diode is reverse biased, entire transformer voltage appears across the diode. The

maximum voltage across the diode is Vm. The diode must be capable to withstand this

voltage. Therefore PIV half wave rating of diode should be equal to Vm in case of single-

phase rectifiers. The average current rating must be greater than Iavg

Full Wave Rectifier:

A single ? phase full wave rectifier using center tap transformer is shown in fig. 10. It supplies

current in both half cycles of the input voltage.

Fig. 10 Fig. 11

In the first half cycle D1 is forward biased and conducts. But D2 is reverse biased and does not

conduct. In the second half cycle D2 is forward biased, and conducts and D1 is reverse

biased. It is also called 2 ? pulse midpoint converter because it supplies current in both the

half cycles. The output voltage waveform is shown in fig. 11.

The average output voltage is given by

and the average load current is given by





When D1 conducts, then full secondary voltage appears across D2, therefore PIV rating of the

diode should be 2 Vm.

Bridge Rectifier:

The single ? phase full wave bridge rectifier is shown in fig. 1. It is the most widely used

rectifier. It also provides currents in both the half cycle of input supply.

Fig. 1 Fig. 2

In the positive half cycle, D1 & D4 are forward biased and D2 & D3 are reverse biased. In the

negative half cycle, D2 & D3 are forward biased, and D1 & D4 are reverse biased. The output

voltage waveform is shown in fig. 2 and it is same as full wave rectifier but the advantage is

that PIV rating of diodes are V m and only single secondary transformer is required.

The main disadvantage is that it requires four diodes. When low dc voltage is required then

secondary voltage is low and diodes drop (1.4V) becomes significant. For low dc output, 2-

pulse center tap rectifier is used because only one diode drop is there.

The ripple factor is the measure of the purity of dc output of a rectifier and is defined as

Therefore,





Clippers:

Clipping circuits are used to select that portion of the input wave which lies above or below

some reference level. Some of the clipper circuits are discussed here. The transfer

characteristic (vo vs vi) and the output voltage waveform for a given input voltage are also

discussed.

Clipper Circuit 1:

The circuit shown in fig. 3, clips the input signal above a

reference voltage (VR).

In this clipper circuit,

If vi < VR, diode is reversed biased and does not

conduct. Therefore, vo = vi

and, if vi > VR, diode is forward biased and thus, vo= VR.

The transfer characteristic of the clippers is shown in fig. 4.

Fig. 3





Clipper Circuit 2:

The clipper circuit shown in fig. 5 clips the input signal

below reference voltage VR.

In this clipper circuit,

If vi > VR, diode is reverse biased. vo = vi

and, If vi < VR, diode is forward biased. vo = VR

The transfer characteristic of the circuit is shown in fig. 6.

Fig. 5

Fig. 6





Clipper Circuit 3:

To clip the input signal between two independent

levels (VR1< VR2 ), the clipper circuit is shown in

fig. 7.

The diodes D1 & D2 are assumed ideal diodes.

For this clipper circuit, when vi ≤ VR1, vo=VR1

and, vi ≥ VR2, vo= VR2

and, VR1 < vi < VR2 vo = vi

The transfer characteristic of the clipper is shown in

fig. 8.

Fig. 7

Fig. 8

Clamper Circuits





Clamping is a process of introducing a dc level into a signal. For example, if the input voltage

swings from -10 V and +10 V, a positive dc clamper, which introduces +10 V in the input will

produce the output that swings ideally from 0 V to +20 V. The complete waveform is lifted up

by +10 V.

Negative Diode clamper:

A negative diode clamper is shown in fig. 8, which introduces a negative dc voltage equal to

peak value of input in the input signal.

Fig. 8 Fig. 9

Let the input signal swings form +10 V to -10

V. During first positive half cycle as V i

rises from 0 to 10 V, the diode conducts.

Assuming an ideal diode, its voltage, which

is also the output must be zero during the

time from 0 to t1. The capacitor charges

during this period to 10 V, with the polarity

shown.

At that Vi starts to drop which means the anode

of D is negative relative to cathode, ( VD =

vi - vc ) thus reverse biasing the diode and

preventing the capacitor from discharging.

Fig. 10




Fig. 9. Since the capacitor is holding its

charge it behaves as a DC voltage source

while the diode appears as an open circuit,

therefore the equivalent circuit becomes an

input supply in series with -10 V dc voltage

as shown in fig. 10, and the resultant output

voltage is the sum of instantaneous input

voltage and dc voltage (-10 V).

Positive Clamper:

The positive clamper circuit is shown in fig. 1, which introduces positive dc voltage equal to

the peak of input signal. The operation of the circuit is same as of negative clamper.

Fig. 1 Fig. 2

Let the input signal swings form +10 V to -10 V. During first negative half cycle as Vi rises

from 0 to -10 V, the diode conducts. Assuming an ideal diode, its voltage, which is also the

output must be zero during the time from 0 to t1. The capacitor charges during this period to

10 V, with the polarity shown.

After that Vi starts to drop which means the anode of D is negative relative to cathode, (VD= vi

- vC) thus reverse biasing the diode and preventing the capacitor from discharging. Fig. 2.

Since the capacitor is holding its charge it behaves as a DC voltage source while the diode

appears as an open circuit, therefore the equivalent circuit becomes an input supply in series







with +10 V dc voltage and the resultant output voltage is the sum of instantaneous input

voltage and dc voltage (+10 V).

To clamp the input signal by a voltage other than peak value, a dc source is required. As shown

in fig. 3, the dc source is reverse biasing the diode.

The input voltage swings from +10 V to -10 V. In the negative half cycle when the voltage

exceed 5V then D conduct. During input voltage variation from ?5 V to -10 V, the capacitor

charges to 5 V with the polarity shown in fig. 3. After that D becomes reverse biased and

open circuited. Then complete ac signal is shifted upward by 5 V. The output waveform is

shown in fig. 4.

Fig. 3 Fig. 4

Voltage Doubler :

A voltage doubler circuit is shown in fig. 5. The circuit produces a dc voltage, which is double

the peak input voltage.







Fig. 5 Fig. 6

At the peak of the negative half cycle D1 is forward based, and D2 is reverse based. This

charges C1 to the peak voltage Vp with the polarity shown. At the peak of the positive half

cycle D1 is reverse biased and D2 is forward biased. Because the source and C1 are in series,

C2 will change toward 2Vp. e.g. Capacitor voltage increases continuously and finally

becomes 20V. The voltage waveform is shown in fig. 6.

To understand the circuit operation, let the input voltage varies from -10 V to +10 V. The

different stages of circuit from 0 to t10 are shown in fig. 7(a).

Fig. 7(a)

During 0 to t1, the input voltage is negative, D1 is forward biased the capacitor is charged to

?10 V with the polarity as shown in fig. 7b.






Expected questions:

1. Derive equation for space charge capacitance for diode

( july/aug 08, jan/feb 10 for 8 M)

2. What is the relation between diode voltage and diode current

(jan/feb 07, jan/feb 08, july/aug 09 for 8

M)

3. What are the effect of temperature on diode operation

( july/aug 09 for 5

M)

4. When a silicon diode is conducting at a temperature of 25°C, a 0.7 V drop exists across

its terminals. What is the voltage, VON, across the diode at 100°C?

( jan/feb 10 for 6 M)

5. Find the output current for the circuit shown in fig.1(a).

( july/aug 09 for 8M)

6. The circuit of fig. 2, has a source voltage of Vs = 1.1 + 0.1 sin 1000t. Find the current,

iD. Assume that

nVT = 40 mV

VON = 0.7 V

( jan/feb 08 for 10M)

7. Explain the small signal operation of diode

( jan/feb 08, july/aug 09 for 8M)

8. Calculate the voltage output of the circuit shown in fig. 5 for following inputs

(a) V1 = V2 = 0.

(b) V1 = V, V2 = 0.

(c) V 1 = V2 = V knew voltage = Vr

Forward resistance of each diode is Rf.







9. Explian the diode application as half wave rectifier Half wave Rectifier

10. ( jan/feb 09, july/aug 08 for 6M)

11. How can a diode circuit be implemented to represent parallel independent clippers

( jan/feb 07, july/aug 09 for 12M)

12. Find the output voltage v out of the clipper circuit of fig. 7(a) assuming that the diodes

are

a. ideal.

b. Von = 0.7 V. For both cases, assume RF is zero.

Fig. 7(a) Fig. 7(b)


13. Using a diode implement voltage doubler circuit

( jan/feb 08 for 8M)

14. Design a 10-volt Zener regulator as shown in fig. 1 for the following conditions:





a. The load current ranges from 100 mA to 200 mA and the source voltage ranges

from 14 V to 20 V. Verify your design using a computer simulation.

b. Repeat the design problem for the following conditions: The load current ranges

from 20 mA to 200 mA and the source voltage ranges from 10.2 V to 14 V.

15. Explain LED operation with an example

Chapter 2. - DC Biasing - BJTs

BJT – A Review

• Invented in 1948 by Bardeen, Brattain and Shockley

• Contains three adjoining, alternately doped semiconductor regions:

Emitter (E), Base (B), and Collector (C)



• The middle region, base, is very thin

• Emitter is heavily doped compared to collector. So, emitter and

collector are not interchangeable.

Three operating regions

• Linear – region operation:

– Base – emitter junction forward biased

– Base – collector junction reverse biased

• Cutoff – region operation:

– Base – emitter junction reverse biased

– Base – collector junction reverse biased

• Saturation – region operation:

– Base – emitter junction forward biased

– Base – collector junction forward biased

Three operating regions of BJT

• Cut off: VCE = VCC, IC ≅ 0

• Active or linear : VCE ≅ VCC/2 , IC ≅ IC max/2

• Saturation: VCE ≅ 0 , IC ≅ IC max

Q-Point

• The intersection of the dc bias value of IB with the dc load line determines the Q- point.

• It is desirable to have the Q-point centered on the load line. Why?

• When a circuit is designed to have a centered Q-point, the amplifier is said to be midpoint

biased.



• Midpoint biasing allows optimum ac operation of the amplifier.

Introduction - Biasing

The analysis or design of a transistor amplifier requires knowledge of both

the dc and ac response of the system.In fact, the amplifier increases the

strength of a weak signal by transferring the energy from the applied DC

source to the weak input ac signal

• The analysis or design of any electronic amplifier therefore has two

components:

• The dc portion and

• The ac portion

During the design stage, the choice of parameters for the required dc

levels will affect the ac response.

What is biasing circuit?

• Once the desired dc current and voltage levels have been identified, a

network must be constructed that will establish the desired values of IB,

IC and VCE, Such a network is known as biasing circuit. A biasing

network has to preferably make

use of one power supply to bias both the junctions of the transistor.

Purpose of the DC biasing circuit

• To turn the device “ON”

• To place it in operation in the region of its characteristic where the device operates most

linearly, i.e. to set up the initial dc values of IB, IC, and VCE

Important basic relationship

• VBE = 0.7V

• IE = (β + 1) IB ≅ IC

• IC = β IB



Collector-to-base bias

This configuration employs negative feedback to prevent thermal runaway and stabilize the

operating point. In this form of biasing, the base resistor RB is connected to the collector

instead of connecting it to the DC source VCC. So any thermal runaway will induce a voltage

drop across the RC resistor that will throttle the transistor's base current.

From Kirchhoff's voltage law, the voltage across the base resistor Rb is

By the Ebers–Moll model, Ic = βIb, and so

From Ohm's law, the base current , and so

Hence, the base current Ib is

http://en.wikipedia.org/wiki/Negative_feedback

http://en.wikipedia.org/wiki/Thermal_runaway

http://en.wikipedia.org/wiki/Kirchhoff%27s_voltage_law

http://en.wikipedia.org/wiki/Ebers%E2%80%93Moll_model

http://en.wikipedia.org/wiki/Ohm%27s_law



RC

If Vbe is held constant and temperature increases, then the collector current Ic increases.

However, a larger Ic causes the voltage drop across resistor Rc to increase, which in turn

reduces the voltage across the base resistor Rb. A lower base-resistor voltage drop reduces

the base current Ib, which results in less collector current Ic. Because an increase in collector

current with temperature is opposed, the operating point is kept stable.

Merits:

Circuit stabilizes the operating point against variations in temperature and β (ie.

replacement of transistor)

Demerits:

In this circuit, to keep Ic independent of β, the following condition must be met:

which is the case when

As β-value is fixed (and generally unknown) for a given transistor, this relation can be

satisfied either by keeping Rc fairly large or making Rb very low.

If Rc is large, a high Vcc is necessary, which increases cost as well as precautions

necessary while handling.

If Rb is low, the reverse bias of the collector–base region is small, which limits

the range of collector voltage swing that leaves the transistor in active mode.

The resistor Rb causes an AC feedback, reducing the voltage gain of the amplifier. This

undesirable effect is a trade-off for greater Q-point stability.

Usage: The feedback also decreases the input impedance of the amplifier as seen from the

base, which can be advantageous. Due to the gain reduction from feedback, this biasing form is

used only when the trade-off for stability is warranted

Biasing circuits:

http://en.wikipedia.org/wiki/Alternating_current

http://en.wikipedia.org/wiki/Voltage_gain

http://en.wikipedia.org/wiki/Q-point



• Fixed – bias circuit

• Emitter bias

• Voltage divider bias

• DC bias with voltage feedback

• Miscellaneous bias

• Applying KVL to the input loop:

VCC = IBRB + VBE

• From the above equation, deriving for IB, we get,

IB = [VCC – VBE] / RB

• The selection of RB sets the level of base current for the operating point.

• Applying KVL for the output loop:

VCC = ICRC + VCE

• In circuits where emitter is grounded,

VCE = VE

VBE = VB

Design and Analysis

• Design: Given – IB, IC , VCE and VCC, or IC , VCE and β, design the

values of RB, RC using the equations obtained by applying KVL to

input and output loops.

• Analysis: Given the circuit values (VCC, RB and RC), determine the values of IB,

IC , VCE using the equations obtained by applying KVL to input and output loops.

Problem – Analysis



Given the fixed bias circuit with VCC = 12V, RB = 240 kΩ, RC = 2.2 kΩ

and β = 75. Determine the values of operating point.

Equation for the input loop is:

IB = [VCC – VBE] / RB where VBE = 0.7V, thus

substituting the other given values in the equation,

we get

IB = 47.08uA

IC = βIB = 3.53mA

VCE = VCC – ICRC = 4.23V

• When the transistor is biased such that IB is very high so as to make IC very high

such that ICRC drop is almost VCC and VCE is almost 0, the transistor is

said to be

in saturation.

IC sat = VCC / RC in a fixed bias circuit.

Load line

The two extreme points on the load line can be calculated and by joining

which the load line can be drawn.

• To find extreme points, first, Ic is made 0 in the equation: VCE = VCC – ICRC .



This gives the coordinates (VCC,0) on the x axis of the output

characteristics.

• The other extreme point is on the y-axis and can be calculated by making VCE = 0

in the equation VCE = VCC – ICRC which gives IC( max) = VCC / RC thus giving the

coordinates of the point as (0, VCC / RC).

• The two extreme points so obtained are joined to form the load line.

• The load line intersects the output characteristics at various points

corresponding to different IBs. The actual operating point is established

for the given IB.

Q point variation

As IB is varied, the Q point shifts accordingly on the load line either up or down depending on

IB increased or decreased respectively.

As RC is varied, the Q point shifts to left or right along the same IB line since the slope of the

line varies. As RC increases, slope reduces ( slope is -1/RC)

which results in shift of Q point to the left meaning no variation in IC and reduction in VCE

.

Thus if the output characteristics is known, the analysis of the given fixed bias circuit

or designing a fixed bias circuit is possible using Emitter Bias

• It can be shown that, including an emitter resistor in the fixed

bias circuit improves the stability of Q point.

• Thus emitter bias is a biasing circuit very

similar to fixed bias circuit with an emitter resistor added to it.

DC bias with voltage

feedback



Input loop

Applying KVL for Input Loop:VCC = IC1RC + IBRB + VBE + IERE

Substituting for IE as (β +1)IB and solving for

IB, IB = ( VCC – VBE) / [ RB + β( RC +

RE)]

Output loop

Neglecting the base current, KVL to the output

loop results in, VCE = VCC – IC ( RC + RE)

Applying KVL to input loop:

VCC = IC|RC + IBRB + VBE + IERE

IC| ≅ IC and IC ≅ IE

Substituting for IE as (β +1)IB [ or as βIB] and solving for

IB, IB = ( VCC – VBE) / [ RB + β( RC + RE)]

Output loop



Neglecting the base current, and applying KVL to the output

loop results in, VCE = VCC – IC ( RC + RE)

In this circuit, improved stability is obtained by introducing a feedback path from

collector to base.

Sensitivity of Q point to changes in beta or temperature variations is

normally less than that encountered for the fixed bias or emitter biased

configurations.

PNP transistors

The analysis of PNP transistors follows the same pattern established for NPN transistors. The

only difference between the resulting equations for a network in which an npn transistor has

been replaced by a pnp transistor is the sign associated with particular quantities.

Applying KVL to Input

loop: VCC = IBRB

+VBE+IERE

Thus, IB = (VCC – VBE) / [RB + (β+1) RE]

Applying KVL Output

loop: VCE = - ( VCC –

ICRC)

Bias stabilization

The stability of a system is a measure of the sensitivity of a network to variations in its parameters.

In any amplifier employing a transistor the collector current IC is sensitive to each of the

following parameters.

β increases with increase in temperature.

Magnitude of VBE decreases about 2.5mV per degree Celsius increase in temperature.

ICO doubles in value for every 10 degree Celsius increase in temperature.



T (degree

Celsius)

Ico (nA) β VBE (V)

- 65

0.2 x 10-3

20

0.85

25

0.1

50

0.65

100

20

80

0.48

175

3.3 x 103

120

0.3

Stability factors

S (ICO) = ∆IC / ∆IC0

S (VBE) = ∆IC / ∆VBE

Networks that are quite stable and relatively insensitive to temperature variations have low

stability factors.

The higher the stability factor, the more sensitive is the network to variations in that parameter.

S( ICO)

• Analyze S( ICO) for

– emitter bias configuration

– fixed bias configuration

– Voltage divider configuration

For the emitter bias configuration,

S( ICO) = ( β + 1) [ 1 + RB / RE] / [( β +

1) + RB / RE] If RB / RE >> ( β + 1)

, then



possible considering all aspects of

S( ICO) = ( β + 1)

For RB / RE <<1, S( ICO)

Thus, emitter bias configuration is quite stable when the ratio RB / RE is as small as possible.

Emitter bias configuration is least stable when RB / RE approaches ( β + 1) .

Fixed bias configuration

S( ICO) = ( β + 1) [ 1 + RB / RE] / [( β + 1) + RB / RE]

= ( β + 1) [RE + RB] / [( β + 1) RE

+ RB] By plugging RE = 0, we

get

S( ICO) = β + 1

This indicates poor stability.

Voltage divider configuration

S( ICO) = ( β + 1) [ 1 + RB / RE] / [( β +

1) + RB / RE] Here, replace RB with

Rth

S( ICO) = ( β + 1) [ 1 + Rth / RE] / [( β + 1) + Rth / RE]

Thus, voltage divider bias configuration is quite stable when the ratio R / R is as small

Physical impact

In a fixed bias circuit, IC increases due to increase in IC0. [IC = βIB + (β+1) IC0]

IB is fixed by VCC and RB. Thus level of IC would continue to rise with temperature –

a very unstable situation.

In emitter bias circuit, as IC increases, IE increases, VE increases. Increase in VE reduces

IB. IB = [VCC – VBE – VE] / RB. A drop in IB reduces IC.Thus, this configuration is

such that there is a reaction to an increase in IC that will tend to oppose the change in bias

conditions.



In the DC bias with voltage feedback, as IC increases, voltage across RC increases, thus

reducing IB and causing IC to reduce.

The most stable configuration is the voltage – divider network. If the condition βRE

>>10R2, the voltage VB will remain fairly constant for changing levels of IC. VBE =

VB – VE, as IC increases, VE increases, since VB is constant, VBE drops making IB to

fall, which will try to offset the increases level of IC.

S(VBE)

S(VBE) = ∆IC / ∆VBE

For an emitter bias circuit, S(VBE) = - β / [ RB + (β + 1)RE]

If RE =0 in the above equation, we get S(VBE) for a fixed bias circuit as, S(VBE) = - β / RB.

For an emitter bias,

S(VBE) = - β / [ RB + (β + 1)RE] can be

rewritten as, S(VBE) = - (β/RE )/

[RB/RE + (β + 1)]

If (β + 1)>> RB/RE, then

The larger the RE, lower the S(VBE) and more stable is

the system. Total effect of all the three parameters on

IC can be written as,

Forward active mode of operation

The forward active mode is obtained by forward-biasing the base-emitter junction. In addition

we eliminate the base-collector junction current by setting VBC = 0. The minority-carrier

distribution in the quasi-neutral regions of the bipolar transistor, as shown in Figure, is used

to analyze this situation in more detail.



Minority-carrier distribution in the quasi-neutral regions of a bipolar transistor (a) Forward

active bias mode. (b) Saturation mode.

The values of the minority carrier densities at the edges of the depletion regions are indicated

on the Figure. The carrier densities vary linearly between the boundary values as expected

when using the assumption that no significant recombination takes place in the quasi-neutral

regions. The minority carrier densities on both sides of the base-collector depletion region

equal the thermal equilibrium values since VBC was set to zero. While this boundary

condition is mathematically equivalent to that of an ideal contact, there is an important

difference. The minority carriers arriving at x = wB - xp,BC do not recombine. Instead, they

drift through the base-collector depletion region and end up as majority carriers in the

collector region.



And the emitter current due to electrons, IE,n, simplifies to:

It is convenient to rewrite the emitter current due to electrons, IE,n, as a function of the total

excess minority charge in the base, Qn,B. This charge is proportional to the triangular area

in the quasi-neutral base as shown in Figure and is calculated from

where tr is the average time the minority carriers spend in the base layer, i.e. the transit time.

The emitter current therefore equals the excess minority carrier charge present in the base

region, divided by the time this charge spends in the base. This and other similar relations

will be used to construct the charge control model of the bipolar junction transistor A

combination of equations yields the transit time as a function of the quasi-neutral layer

width, wB', and the electron diffusion constant in the base, Dn,B.

We now turn our attention to the recombination current in the quasi-neutral base and obtain it

from the continuity equation

By applying it to the quasi-neutral base region and assuming steady state conditions:



which in turn can be written as a function of the excess minority carrier charge, Qn,B, using

equation

It is typically the emitter efficiency, which limits the current gain in transistors made of silicon

or germanium. The long minority-carrier lifetime and the long diffusion lengths in those

materials justify the exclusion of recombination in the base or the depletion layer. The

resulting current gain, under such conditions, is:

From this equation, we conclude that the current gain can be larger than one if the emitter

doping is much larger than the base doping. A typical current gain for a silicon bipolar

transistor is 50 - 150.

Expected Questions:

1. What are the operating regions of BJT

2. Explain the biasing circuits of BJT

3. Explain load line analysis of BJT



4Analyze the following circuit: given

β = 75, VCC = 16V, RB = 430kΩ, RC = 2kΩ and RE = 1k Ω

1. How to achieve improved bias stability

2. For the circuit given below, find IC and VCE.

Given the values of R1, R2, RC, RE and β = 140 and VCC =

18V.

For the purpose of DC analysis, all the capacitors in the

amplifier circuit are opened.

3. Given: IB = ( VCC – VBE) / [ RB + β( RC + RE)]= 11.91µA

IC = (β IB ) = 1.07mA

VCE = VCC – IC ( RC + RE) = 3.69V

4. Determine the DC level of IB and VC for the network shown:



5. Analyze the circuit in the next slide. Given β = 120

6. Find VCB and IB for the Common base configuration given:

Given: β = 60

7. The emitter bias circuit has the following specifications: ICQ = 1/2Isat, Isat = 8mA,

VC =

18V, VCC = 18V and β = 110. Determine RC , RE and RB.

Unit-3

Uni-junction transistor

Uni-junction transistor

The UJT as the name implies, is characterized by a single pn junction. It exhibits negative

resistance characteristic that makes it useful in oscillator circuits.

The symbol for UJT is shown in fig. 1. The UJT is having three terminals base1 (B1), base2

(B2) and emitter (E). The UJT is made up of an N-type silicon bar which acts as the base as

shown in fig. 2. It is very lightly doped. A P-type impurity is introduced into the base,

producing a single PN junction called emitter. The PN junction exhibits the properties of a

conventional diode.





Fig. 1

Fig .2

A complementary UJT is formed by a P-type base and N-type emitter. Except for the polarity

of voltage and current the characteristic is similar to those of a conventional UJT.

A simplified equivalent circuit for the UJT is shown in fig. 3. VBB is a source of biasing

voltage connected between B2 and B1. When the emitter is open, the total resistance from

B2 to B1 is simply the resistance of the silicon bar, this is known as the inter base resistance

RBB. Since the N-channel is lightly doped, therefore RBB is relatively high, typically 5 to

10K ohm. RB2 is the resistance between B2 and point ?a', while RB1 is the resistance from

point ?a' to B1, therefore the interbase resistance RBB is

RBB = RB1 + RB2




Fig. 3

The diode accounts for the rectifying properties of the PN junction. VD is the diode's threshold

voltage. With the emitter open, IE = 0, and I1 = I 2 . The interbase current is given by

I1 = I2 = VBB / R BB .

Part of VBB is dropped across RB2 while the rest of voltage is dropped across RB1. The voltage

across RB1 is

Va = VBB * (RB1 ) / (RB1 + RB2 )

The ratio RB1 / (RB1 + RB2 ) is called intrinsic standoff ratio

= RB1 / (RB1 + RB2 ) i.e. Va = VBB .

The ratio is a property of UJT and it is always less than one and usually between 0.4 and

0.85. As long as IB = 0, the circuit of behaves as a voltage divider.

Assume now that vE is gradually increased from zero using an emitter supply VEE . The diode

remains reverse biased till vE voltage is less than VBB and no emitter current flows except

leakage current. The emitter diode will be reversed biased.

When vE = VD + VBB, then appreciable emitter current begins to flow where VD is the diode's

threshold voltage. The value of vE that causes, the diode to start conducting is called the

peak point voltage and the current is called peak point current IP.

VP = VD + VBB.

The graph of fig. 4 shows the relationship between the emitter voltage and current. vE is plotted

on the vertical axis and IE is plotted on the horizontal axis. The region from vE = 0 to vE =

VP is called cut off region because no emitter current flows (except for leakage). Once vE

exceeds the peak point voltage, IE increases, but v E decreases. up to certain point called

valley point (VV and IV). This is called negative resistance region. Beyond this, IE increases

with vE this is the saturation region, which exhibits a positive resistance characteristic.

The physical process responsible for the negative resistance characteristic is called

conductivity modulation. When the vE exceeds VP voltage, holes from P emitter are injected




into N base. Since the P region is heavily doped compared with the N-region, holes are

injected to the lower half of the UJT

The lightly doped N region gives these holes a long lifetime. These holes move towards B1 to

complete their path by re-entering at the negative terminal of VEE. The large holes create a

conducting path between the emitter and the lower base. These increased charge carriers

represent a decrease in resistance RB1, therefore can be considered as variable resistance. It

decreases up to 50 ohm.

Since is a function of RB1 it follows that the reduction of RB1 causes a corresponding

reduction in intrinsic standoff ratio. Thus as IE increases, RB1 decreases, decreases, and Va

decreases. The decrease in V a causes more emitter current to flow which causes further

reduction in RB1, , and Va. This process is regenerative and therefore Va as well as vE

quickly drops while IE increases. Although RB decreases in value, but it is always positive

resistance. It is only the dynamic resistance between VV and VP. At point B, the entire base1

region will saturate with carriers and resistance RB1 will not decrease any more. A further

increase in Ie will be followed by a voltage rise.

The diode threshold voltage decreases with temperature and RBB resistance increases with

temperature because Si has positive temperature coefficient.

UJT Relaxation Oscillator:

The characteristic of UJT was discussed in previous lecture. It is having negative resistance

region. The negative dynamic resistance region of UJT can be used to realize an oscillator.

The circuit of UJT relaxation oscillator is shown in fig. 1. It includes two resistors R1 and R2

for taking two outputs R2 may be a few hundred ohms and R1 should be less than 50 ohms.




The dc source VCC supplies the necessary bias. The interbase voltage VBB is the difference

between VCC and the voltage drops across R1 and R2. Usually RBB is much larger than R1

and R2 so that VBB approximately equal to V. Note, RB1 and RB2 are inter-resistance of UJT

while R1 and R2 is the actual resistor. RB1 is in series with R1 and RB2 is in series with R2 .

As soon as power is applied to the circuit capacitor begins to charge toward V. The voltage

across C, which is also VE , rises exponentially with a time constant

As long as VE < VP, IE = 0. the diode remains reverse biased as long as VE < VP . When the

capacitor charges up to VP , the diode conducts and RB1 decreases and capacitor starts

discharging. The reduction in R B1 causes capacitor C voltage to drop very quickly to the

valley voltage VV because of the fast time constant due to the low value of RB1 and R1. As

soon as VE drops below Va + VD the diode is no longer forward biased and it stops

conduction. It now reverts to the previous state and C begins to charge once more toward

VCC .

The emitter voltage is shown in fig. 2, VE rises exponentially toward VCC but drops to a very

low value after it reaches VP. The time for the VE to drop from VP to VV is relatively small

and usually neglected. The period T can therefore be approximated as follows:




There are two additional outputs possible for the UJT oscillation one of these is the voltage

developed at B1 due to capacitor discharge while the other is voltage developed at B2 as

shown in fig. 3.

When UJT fires (at t = T) Va drops, causing a

corresponding voltage drop at B2. The duration of

outputs at B1 and B2 are determined by C discharge

time.

If R1 is very small, C discharges very quickly and very

narrow pulse is produced at the output. If R1 = 0,

obviously no pulses appear at B1.

If R2 = 0, no pulse can be generated at B2. If R1 is too

large, its positive resistance may swamp the negative

resistance and prevent the UJT form switching back

after it has fired.

R2, in addition to providing a source of pulse at B2, is

useful for temperature stabilization of the UJT's peak

point voltage .

VP = VD + VBB.

As the temperature increases, Vp decreases. The

temperature coefficient of RBB is positive. Rs is

essentially independent of temperature. It is therefore

possible to select R2 so that VBB increases with

temperature by the same amount as VD decreases.

This provides a constant VP and, in turn, frequency

of oscillation.

Figure 31.3

Selection of R and C:

In the circuit, R is required to pass only the capacitor charging current. At the instant when VP

is reached; R must supply the peak current. It is therefore, necessary, that the current

through R should be slightly greater than the peak point.




Once the UJT fires, VE drops to the valley voltage VV . IE should not be allowed to increases

beyond the valley point IV, otherwise the UJT is taken into saturation region and does not

switch back, R therefore must be selected large enough to ensure that

As long as R is chosen between these extremes, reliable operation results.

Expected Questions

1. What is the need for bias compensation?

2. Explain the compensation techniques used for Vbe and Icbo

3. What do you mean by rhermal runway of a transistor? Explain

4. Define h-parameter

5. Draw the complete hybrid equivalent circuit of a transistor

6. Sketch the re- equivalent circuit for Ce fixed bias configuration and derive the

expression for Ar, Ai,Zi and Zo

7. What is bias stabilization? Explain

8. Derive an exzpression for Sico and SVbe for Fixed bias configuration



UNIT-4

Frequency Response of Amplifier

Frequency curve of an RC coupled amplifier:

A practical amplifier circuit is meant to raise the voltage level of the input signal. This signal

may be obtained from anywhere e.g. radio or TV receiver circuit. Such a signal is not of a

single frequency. But it consists of a band of frequencies, e.g. from 20 Hz to 20 KHz. If the

loudspeakers are to reproduce the sound faithfully, the amplifier used must amplify all the

frequency components of signal by same amount. If it does not do so, the output of the

loudspeaker will not be the exact replica of the original sound. When this happen then it

means distortion has been introduced by the amplifier. Consider an RC coupled amplifier

circuit shown in fig. 1.

Fig. 1 Fig. 2

Fig. 2, shows frequency response curve of a RC coupled amplifier. The curve is usually plotted

on a semilog graph paper with frequency range on logarithmic scale so that large frequency

range can be accommodated. The gain is constant for a limited band of frequencies. This

range is called mid-frequency band and gain is called mid band gain. AVM. On both sides of

the mid frequency range, the gain decreases. For very low and very high frequencies the

gain is almost zero.





In mid band frequency range, the coupling capacitors and bypass capacitors are as good as

short circuits. But when the frequency is low. These capacitors can no longer be replaced by

the short circuit approximation.

First consider coupling capacitor. The ac equivalent is shown in fig. 3, assuming capacitors are

offering some impedance. In mid-frequency band, the capacitors are ac shorted so the input

voltage appears directly acrossr'e but at low frequency the XC is significant and some

voltage drops across XC. The input vin at the base decreases. Thus decreasing output voltage.

The lower the frequency the more will be XC and lesser will be the output voltage.

Fig. 3

Similarly at low frequency, output capacitor reactance also increases. The voltage across RL

also reduces because some voltage drop takes place across XC. Thus output voltage reduces.

The XC reactance not only reduces the gain but also change the phase between input and

output. It would not be exactly 180o but decided by the reactance. At zero frequency, the

capacitors are open circuited therefore output voltage reduces to zero.




The other component due to which gain decreases at low frequencies is the bypass capacitor.

The function of this capacitor is to bypass ac and blocks dc The impedence of this capacitor

in mid frequency band is very low as compared to RE so it behaves like ac short but as the

frequency decrease the XCE becomes more and no longer behaves like ac short. Now the

emitter is not ac grounded. The ac emitter current i.e. divides into two parts i1 and i2, as

shown in fig. 4. A current i1 passes through RE and rest of the current passes through C. Due

to ac current i1 in RE, an ac voltage is developed i1 * RE. With the polarity marked at an

instant. Thus the effective VL voltage is given by

Vbe = Vs ? RE.

Thus the effective voltage input is reduced. The output also reduces. The lower the frequency,

the lesser will be the gain. This reduction in gain is due to negative feedback.

As the frequency of the input signal increases, again the gain of the amplifier reduces. Firstly

the of the transistor decreases at higher frequency. Thus reducing the voltage gain of the

amplifier at higher frequencies as shown in fig. 5.

The other factor responsible for the reduction in gain at higher frequencies is the presence of

various capacitors as shown in fig. 6. They are not physically connected but inherently

present with the device.






Fig. 5

The capacitor Cbc between the base and the collector connects the output with the input.

Because of this, negative feedback takes place in the circuit and the gain decreases. This

feedback effect is more, when Cbc provides a path for higher frequency ac currents

The capacitance Cbe offers a low input impedance at higher frequency thus reduces the

effective input signal and so the gain falls. Similarly, Cce provides a shunting effect at high

frequencies in the output side and reduces gain of the amplifier.

Besides these junction capacitances there are wiring capacitance CW1 and CW2. These reactance

are very small but at high frequencies they become 5 to 20 p.f. For a multistage amplifier,

the effect of the capacitances Cce,CW1 and CW2 can be represented by single shunt

capacitance.

CS = CW1 + CW2 +Cce.

At higher frequency, the capacitor CS offers low input impedance and thus reduces the output.

Fig. 6

Bandwidth of an amplifier:



The gain is constant over a frequency range. The frequencies at which the gain reduces to

70.7% of the maximum gain are known as cut off frequencies, upper cut off and lower cut

off frequency. fig. 7, shows these two frequences. The difference of these two frequencies is

called Band width (BW) of an amplifier.

BW = f2 ? f1.

Fig. 7

At f1 and f2, the voltage gain becomes 0.707 Am(1 / 2). The output voltage reduces to 1 /

2 of maximum output voltage. Since the power is proportional to voltage square, the output

power at these frequencies becomes half of maximum power. The gain on dB scale is given

by

20 log10(V2 / V1) = 10 log 10 (V2 / V1)2 = 3 dB.

20 log10(V2 / V1) = 20 log10(0.707) =10 log10 (1 / 2)2 = 10 log10(1 / 2) = -3 dB.

If the difference in gain is more than 3 dB, then it can be detected by human. If it is less than 3

dB it cannot be detected.

Direct Coupling:

For applications, where the signal frequency is below 10 Hz, coupling and bypass capacitors

cannot be used. At low frequencies, these capacitors can no longer be treated as ac short

circuits, since they offer very high impedance. If these capacitors are used then their values

have to be extremely large e.g. to bypass a 100 ohm emitter resistor at 10 Hz, we need a




capacitor of approximately 1600 F. The lower the frequency the worse the problem

becomes.

To avoid this, direct coupling is used. This means designing the stages without coupling and

bypass capacitors, so that the direct current is coupled as well as alternating current. As a

result, there is no lower frequency limit. The amplifier enlarges the signal no matter have

low frequency including dc or zero frequency.

One Supply Circuit:

Fig. 8, shows a two stage direct coupled amplifier, no coupling or bypass capacitors are used.

With a quiescent input voltage 1.4 V, emitter voltage = 1.4 - 0.7 = 0.7 V




Fig. 8

The output varies from +6V to +8V.

The main disadvantage is variation in transistor characteristic with variation in temperature.

This causes IC and VC to change. Because of the direct coupling the voltage changes are

coupled from one stage to next stage, appearing at the final output as an amplified voltage.

The unwanted change is called drift.

Grounded Reference Input

For the above amplifier, we need a quiescent voltage of 1.4V. In most applications, it is

necessary to have grounded reference input one where the quiescent input voltage is 0 V, as

shown in fig. 9.




Fig. 9

The quiescent V CE of the first transistor is only 0.7V and the quiescent of the second transistor

is only 1.4V. Both the transistors are operating in active region because VCE(sat) is only 0.1

volt. The input is only in mV, which means that these transistors continue to operate in the

active region when a small signal is present.

h-Parameters

Small signal low frequency transistor Models:

All the transistor amplifiers are two port networks having two voltages and two currents. The

positive directions of voltages and currents are shown in fig. 1.

Fig. 1




Out of four quantities two are independent and two are dependent. If the input current i1 and

output voltage v2 are taken independent then other two quantities i2 and v1 can be expressed

in terms of i1 and V2.

The equations can be written as

where h11, h12, h21 and h22 are called h-parameters.

= hi = input impedance with output short circuit to ac.

=hr = fraction of output voltage at input with input open circuited or reverse voltage gain with

input open circuited to ac (dimensions).

= hf = negative of current gain with output short circuited to ac.

The current entering the load is negative of I2. This is also known as forward short circuit

current gain.

= ho = output admittance with input open circuited to ac.



If these parameters are specified for a particular configuration, then suffixes e,b or c are also

included, e.g. hfe ,h ib are h parameters of common emitter and common collector amplifiers

Using two equations the generalized model of the amplifier can be drawn as shown in fig. 2.

Fig. 2

The hybrid model for a transistor amplifier can be derived as follow:

Let us consider CE configuration as show in fig. 3. The variables, iB, iC ,vC, and vB represent

total instantaneous currents and voltages iB and vC can be taken as independent variables and

vB, IC as dependent variables.





Fig. 3

vB = f1 (iB ,vC )

IC = f2 ( iB , vC ).

Using Taylor 's series expression, and neglecting higher order terms we obtain.

The partial derivatives are taken keeping the collector voltage or base current constant. The Δ

vB, Δ vC, Δ iB, Δ iC represent the small signal (incremental) base and collector current and

voltage and can be represented as vb ,ib ,vC ,iC.

The model for CE configuration is shown in fig. 4.

Fig. 4




Expected questions:

1. Obtain an expression in term of h parameter for a transistor as a two-port

network.

2. Using the above developed equation obtain the hybrid model of CE, CC and CB

configuration

3. State and explain Millers theorem

4. Derive an equation for millers input and output capacitance

5. Discuss low frequency response if BJT amplifier and give expression for low

cut-off frequency due to Co, Ce, Cs

UNIT-5

General Amplifiers

Darlington Amplifier:

It consists of two emitter followers in cascaded mode as shown in fig. 1. The overall gain is

close to unity. The main advantage of Darlington amplifier is very large increase in input

impedence and an equal decrease in output impedance .




DC Analysis:

The first transistor has one VBE drop and second transistor has second VBE drop. The voltage

divider produces VTH to the input base. The dc emitter current of the second stage is

IE2 = (VTH ? 2 vBE ) / (RE )

The dc emitter current of the first stage that is the base current of second stage is given by

If r'e(2) is neglected then input impedance of second stage is

This is the impedance seen by the first transistor. If r'e(1) is also neglected then the input

impedance of 1 becomes.

which is extremely high because of the products of two betas, so the approximate input

impedance of Darlington amplifier is

Zin = R1 || R2

Output impedance:

The Thevenin impedance at the input is given by

RTH = RS || R1 || R2

Similar to single stage common collector amplifier, the output impedance of the two stages

zout(1) and zout(2) are given by.

Therefore, t he output impedance of the amplifier is very small.





Zin = R1 || R2

Output impedance:


RTH = RS || R1 || R2






Zin = R1 || R2

Output impedance:


RTH = RS || R1 || R2





Therefore, t he output impedance of the amplifier is very small. Output impedance:


RTH = RS || R1 || R2




Efficiency is given by

Value is around 40%

Power rating is an important consideration in selecting bias resistors since they must be

capable of withstanding the maximum anticipated (worst case) power without overheating.

Power considerations also affect transistor selection. Designers normally select components

having the lowest power handling capability suitable for the design. Frequently, de-rating

(i.e., providing a "safety margin" from derived values) is used to improve the reliability of a



device. This is similar to using safety factors in the design of mechanical systems where the

system is designed to withstand values that exceed the maximum.

Consider a common emitter amplifier circuit shown in fig. 1.

Fig. 1

Derivation of Power Equations

Average power is calculated as follows:

For dc: (E-1)

For ac: (E-2)

In the ac equation, we assume periodic waveforms where T is the period. If the signal is not

periodic, we must let T approach infinity in equation E-1. Looking at the CE amplifier of

fig. 1, the power supplied by the power source is dissipated either in R1 and R2 or in the

transistor (and its associated collector and emitter circuitry). The power in R1 and R2 (the

bias circuitry) is given by

(E-3)





where IR1 and IR2 are the (downward) currents in the two resistors. Kirchhoff's current law

(KCL) yields a relationship between these two currents and the base quiescent current.

IR1 = IR2 ? IB (E-4)

KVL yields the base loop equation (assuming VEE = 0),

IR2 R2 + IR1 R1 = VCC (E-5)

These two equations can be solved for the currents to yield,

(E-6)

In most practical circuits, the power due to IB is negligible relative to the power dissipated in

the transistor and in R1 and R2. We will therefore assume that the power supplied by the

source is approximately equal to the power dissipated in the transistor and in R1 and R2.

This quantity is given by

(E-7)

Where the source voltage VCC is a constant value. The source current has a dc quiescent

component designated by iCEQ and the ac component is designated by ic(t). The last equality

of Equation (E-7) assumes that the average value of ic(t) is zero. This is a reasonable

assumption. For example, it applies if the input ac signal is a sinusoidal waveform.

The average power dissipated by the transistor itself (not including any external circuitry) is

(E-8)

For zero signal input, this becomes

P(transitor) = VCEQ ICQ

Where VCEQ and ICQ are the quiescent (dc) values of the voltage and current, respectively.



For an input signal with maximum possible swing (i.e., Q-point in middle and operating to

cutoff and saturation),

Fig. 2

Putting these time functions in Equation (E-7) yields the power equation,

(E-10)

From the above derivation, we see that the transistor dissipates its maximum power (worst

case) when no ac signal input is applied. This is shown in fig. 2, where we note that the

frequency of the instantaneous power sinusoid is 2ω.

Depending on the amplitude of the input signal, the transistor will dissipate an average power

between VCEQ ICQ and one half of this value. Therefore, the transistor is selected for zero

input signal so it will handle the maximum (worst case) power dissipation of VCEQ ICQ.

We will need a measure of efficiency to determine how much of the power delivered by the

source appears as signal power at the output. We define conversion efficiency as




Cascade Amplifier:

To increases the voltage gain of the amplifier, multiple amplifier are connects in cascade. The

output of one amplifier is the input to another stage. In this way the overall voltage gain can

be increased, when number of amplifier stages are used in succession it is called a

multistage amplifier or cascade amplifier. The load on the first amplifier is the input

resistance of the second amplifier. The various stages need not have the same voltage and

current gain. In practice, the earlier stages are often voltage amplifiers and the last one or

two stages are current amplifiers. The voltage amplifier stages assure that the current stages

have the proper input swing. The amount of gain in a stage is determined by the load on the

amplifier stage, which is governed by the input resistance to the next stage. Therefore, in

designing or analyzing multistage amplifier, we start at the output and proceed toward the

input.

A n-stage amplifier can be represented by the block diagram as shown in fig. 3.

Fig. 3

In fig. 3, the overall voltage gain is the product of the voltage gain of each stage. That is, the

overall voltage gain is ABC.





To represent the gain of the cascade amplifier, the voltage gains are represents in dB. The two

power levels of input and output of an amplifier are compared on a logarithmic scale rather

than linear scale. The number of bels by which the output power P2 exceeds the input power

P1 is defined as

Because of dB scale the gain can be directly added when a number of stages are cascaded.

Types of Coupling:

In a multistage amplifier the output of one stage makes the input of the next stage. Normally a

network is used between two stages so that a minimum loss of voltage occurs when the

signal passes through this network to the next stage. Also the dc voltage at the output of one

stage should not be permitted to go to the input of the next. Otherwise, the biasing of the

next stage are disturbed.

The three couplings generally used are.

1. RC coupling

2. Impedance coupling

3. Transformer coupling.



1.RC coupling:

Fig. 4 shows RC coupling the most commonly used method of coupling from one stage to the

next. An ac source with a source resistance R S drives the input of an amplifier. The

grounded emitter stage amplifies the signal, which is then coupled to next CE stage the

signal is further amplified to get larger output.

In this case the signal developed across the collector resistor of each stage is coupled into the

base of the next stage. The cascaded stages amplify the signal and the overall gain equals

the product of the individual gains.

Fig. 4

The coupling capacitors pass ac but block dc Because of this the stages are isolated as for as dc

is concerned. This is necessary to avoid shifting of Q-points. The drawback of this approach

is the lower frequency limit imposed by the coupling capacitor.

The bypass capacitors are needed because they bypass the emitters to ground. Without them,

the voltage gain of each stage would be lost. These bypass capacitors also place a lower

limit on the frequency response. As the frequency keeps decreasing, a point is reached at

which capacitors no longer look like a.c. shorts. At this frequency the voltage gain starts to

decrease because of the local feedback and the overall gain of the amplifier drops

significantly. These amplifiers are suitable for frequencies above 10 Hz.

Example - 1




Determine the current and voltage gains for the two-stage capacitor-coupled amplifier shown

in fig. 1.

Fig. 1

Solution:

We develop the hybrid equivalent circuit for the multistage amplifier. This equivalent is shown

in fig. 2. Primed variables denote output stage quantities and unprimed variables denote

input stage quantities.

Fig. 2

Calculations for the output stages are as follows





For the input stage,

The input resistance is determined as:

The current gain, Ai, can be found by applying the equations derived earlier, where the first

stage requires using the correct value for Rload derived form the value of Rin to the next

stage.

Alternatively, we analyze fig. 2 by extracting four current dividers as shown in fig. 3.





Fig. 3

The current division of the input stage is

The output of the first stage is coupled to the input of he second stage in fig. 3(b). The input

resistance of the second stage is

The current in R'in is iload and is given by

Again, i load is current-divided at the input to the second stage. Thus,

The output current is found from fig. 3(c):





The current gain is then

Ai =927

Now using the gain impedance formula, we find the voltage gain:

Impedance Coupling:

At higher frequency impedance coupling is used. The collector resistance is replaced by an

inductor as shown in fig. 4. As the frequency increases, XL approaches infinity and each

inductor appears open. In other words, inductors pass dc but block ac. When used in this

way, the inductors are called RFchokes.

Fig. 4

The advantage is that no signal power is wasted in collector resistors. These RF chokes are

relatively expensive and their impedance drops off at lower frequencies. It is suitable at

radio frequency above 20 KHz.

Transformer Coupling:

In this case a transformer is used to transfer the ac output voltage of the first stage to the input

of the second stage. Fig. 5, the resistors RC is replaced by the primary winding of the





transformer. The secondary winding is used to give input to next stage. There is no coupling

capacitor. The dc isolation between the two stages provided by the transformer itself. There

is no power loss in primary winding because of low resistance.

Fig. 5

At low frequency the size and cost of the transformer increases. Transformer coupling is still

used in RF amplifiers. In AM radio receivers, RF signal have frequencies 550 to 1600 KHz.

In TV receivers, the frequencies are 54 to 216 MHz. At these frequency the size and cost of

the transformer reduces. CS capacitor is used to make other point of transformer grounded,

so that ac signal is applied between base and ground.

Tuned Transformer Coupling:

In this case a capacitor is shunted across primary winding to get resonance as shown in fig. 6.

At this frequency the gain is maximum and at other frequencies the gain reduces very much.

This allows us to filter out all frequencies except the resonant frequency and those near it.

This is the principle behind tuning in a radio station or TV channel.




Fig. 6

Example - 2

Design a transformer-coupled amplifier as shown in fig. 7 for a current gain of Ai = 80. Find

the power supplied to the load and the power required from the supply.

Fig. 7

Solution:

We first use the design equation to find the location of the Q-point for maximum output swing.




Since the problem statement requires a current gain of 80, the amplifier must have a current

gain of 10 because the transformer provides an additional gain of 8. We use the equations

from Chapter 5 to find the base resistance RB,

We note that re is sufficiently small to be neglected. Then, solving for RB yields

Now solving for the bias resistors,

The design is now complete. The power delivered by the source is given by

The power dissipated in the load is

We have restricted operation to the linear region by eliminating 5% of the maximum swing

near cutoff and saturation. The efficiency is the ratio of the load to source power.



Expected Questions:

1. Explain BJT relaxation oscillator

2. How is UJT formed

3. Draw the cascade configuration and list the advantages of this circuit

4. With necessary equavalint circuit diagram obtain the expression for Zin Zo and Av for

darlington emitter follower

5. Derive xpression for Zif and Zof for volatages series feed back amplifier and list the

advantages of negative feed back amplifier

6. With block diagram explain the difference between voltage series and voltage shunt

feedback

7. Discuss the general characteristics os a negative feedback amplifier

8. Give classification of multistage amplifier

9. Explain the various distortion in amplifiers.



UNIT-6

Power Amplifier

The amplifiers in multistage amplifier near the load end in almost all-electronic system employ

large signal amplifiers (Power amplifiers) and the purpose of these amplifiers is to obtain

power again.

Consider the case of radio receiver, the purpose of a radio receiver is to produce the transmitted

programmes with sufficient loudness. Since the radio signal received at the receiver output

is of very low power, therefore, power amplifiers are used to put sufficient power into the

signal. But these amplifier need large voltage input.

Therefore, it is necessary to amplify the magnitude of input signal by means of small

amplifiers to a level that is sufficient to drive the power amplifier stages.

In multistage amplifier, the emphasis is on power gain in amplifier near the load. In these

amplifies, the collector currents are much larger because the load resistances are small (i.e

impedence of loud speaker is 3.2 ohm).

A power amplifier draws a large amount of dc power form dc source and convert it into signal

power. Thus, a power amplifier does not truly amplify the signal power but converts the dc

power into signal power.

DC and AC load lines:

Consider a CE amplifier as shown in fig.



The dc equivalent circuit gives the dc load line as shown in fig. 2.

Fig. 2

Q is the operative point. ICQ and V CEQ are quiescent current and voltage. The ac equivalent

circuit is shown in fig. 3.

Fig. 3

This circuit produces ac load line. When no signal is present, the transistor operates at the Q

point shown in fig. 4.






Fig. 4

When a signal is present, operating point swings along the ac load line rather than dc load line.

The saturation and cut off points on the ac load line are different from those on the dc load

line.

During the positive half cycle of ac source, voltage, the collector voltage swing from the Q-

point towards saturation. On the negative cycle, the collector voltage swings from Q-point

towards cutoff. For a large signal clipping can occur on either side or both sides.

The maximum positive swing from the Q-point is



The maximum negative swing from the Q-point is

The ac output compliance (maximum peak to peak unclipped voltage) is given by the smaller

of these two approximate values:

PP = 2 I CQ rC

or PP = 2 V CEQ.

Class A operation:

In a class A' operation transistor operates in active region at all times. This implies that

collector current flows for 360° of the ac cycle.

Voltage gain of loaded amplifier

Current gain

ac input power to the base Pin = V in i b

ac output power point = - V out * i C. ( Negative sign is due to phase inversion.)



The variation of PL with VPP is shown in fig. 5. Maximum ac load power is obtained when the

output unclipped voltage equals ac output compliance PP.

Fig. 5

When no signal drives the amplifier, the power dissipation of the transistor equals the product

of d. voltage and current

P DQ = V CEQ * I CQ

When there is no input signal, PD is maximum as shown in fig. 6.





Fig. 6

It decreases when the peak to peak load voltage increases. The power dissipation must be less

than the rating of transistor, otherwise temperature increases and transistor may damage. To

reduce the temperature, heat sinks are used that dissipates the heat produced. When Q-point

is at the center of ac load line then peak swing above and below Q-point is equal.

Class A current drain:

In a class A amplifier shown in fig.1, the dc source VCC must supply direct current to the

voltage divider and the collector circuit.




Fig. 1

Assuming a stiff voltage divider circuit, the dc current drain of the voltage divider circuit is

I 1 = V CC / (R 1 +R 2 )

In the collector circuit, the dc current drain is

I 2 = I CQ

In a class A amplifier, the sinusoidal variations in collector current averages to zero. Therefore,

whether the ac signal is present or not, the dc source must supply an average current of

I S = I 1 + I 2.

This is the total dc current drain. The dc source voltage multiplied by the dc current drain gives

the ac power supplied to an amplifier.

P S = V CC IS

Therefore, efficiency of the amplifier, = (P L (max) / P S ) * 100 %

Where,, P L (max) = maximum ac load line power. In class A amplifier, there is a wastage of

power in resistor R C and R E i.e. ICQ 2 * (R C + R E ).



To reduce this wastage of power R C and R E should be made zero. R E cannot be made zero

because this will give rise to bias stability problem. R C can also not be made zero because

effective load resistance gets shorted. This results in more current and no power transfer to

the load R L. The R C resistance can, however, be replaced by an inductance whose dc

resistance is zero and there is no dc voltage drop across the choke as shown in fig. 1.

Since in most application the load is loudspeaker, therefore power amplifier drives the

loudspeaker, and the maximum power transfer takes place only when load impedence is

equal to the source impedence. If it is not, the loud speaker gets less power. The impedence

matching is done with the help of transformer, as shown in fig. 2.

Fig. 2

The ratio of number of turns is so selected that the impedence referred to primary side can be

matched with the output impedence of the amplifier.





Class B amplifier:

The efficiency ( ) of class A amplifier is poor. The reason is that these circuits draw

considerable current from the supply even in the absence of input signals.

In class B operation the transistor collector current flows for only 180° of the ac cycle. This

implies that the Q-point is located approximately at cutoff on both dc and ac load lines. The

advantages of class B operation are

Lower transistor power dissipation

Reduced current drain.

Eficiency is given by

Value is around 70%

Biasing a class B amplifier:

In class B amplifier, two complement any transistors are required. Because of the series

connection, each transistor drops half the supply voltage. To avoid cross over distortion, the

Q-point slightly above cut off, with the correct VBE somewhere between 0.6 and 0.7.

If there is an increase in VBE by few mV it produces 10 times as much emitter current. Because

of this it is difficult to find standard resistors that can produce the correct VBE and it needs

an adjustable resistor.

The biasing does not solve thermal instability problem. Because for a given collector current,

VBE requirement decreases by 2 mV per degree rise in temperature. The voltage divider

produces a stiff drive for each diode. Therefore as the temperature increases, the fixed

voltage on each emitter diode forces the collector current to increase and this gives rise to

thermal run away. When the temperature increases collector current increases, and this is

equivalent to Q-point moving up along the vertical dc load line. As the Q-point moves

toward higher collector currents, the temperature of the transistor increases further reducing

the required V BE .



Fig. 3

One way to avoid thermal run away is to use diode bias. It is based on the concept of current

mirror as shown in fig. 3, the base current is much smaller than the current through the

resistor and diode. For this reason, I1 and I2 are approximately equal. If the diode curve is

identical to the VBE curve of the transistor (VBE , IE ). The diode current equals the emitter

and also collector current. Therefore I1 is nearly equal to IC.

I1 = I C .

The collector current is set by controlling the resistor current. This is called a current mirror.

Similarly, pnp transistor can be used as a current mirror. If the VBE curve of the transistor

matches the diode curve, the collector equals the resistor current.

Diode bias of class B push pull emitter follower relies on two current mirrors as shown in fig.

4.







The upper half is an npn current mirror, and the lower half is a pnp current mirror as shown

in fig. 4. For diode bias to be immune to changes in temperature, the diode curve must

match the VBE curves of the transistor over a wide temperature range. This is easily done in

ICs .

Power Calculations for Class B Push-Pull Amplifier

The power delivered by the ac source is split between the transistor and the resistors in the bias

circuitry. The ac signal source adds an insignificant additional amunt of power since base

currents are small relative to collector currents. Part of the power to the transistor goes to

load, and the other part is dissipated by the transistor itself. The following equations specify

the various power relationships in the circuit.

The average power supplied by the dc source is

iCC(t) is the total current and is composed of two components: the dc current through the base

bias resistor and diode combination, and the ac collector current through transistor, Q1.

Under quiescent conditions (i.e. zero input) Q1 is in cutoff mode. Collector current flows

during the positive half of the output signal waveform. Therefore we only need to integrate

this component of the power supply signal over the first half cycle.

The maximum values of collector current and power delivered to the transistor are

The ac output power, assuming a sinusoidal input, is




The maximum ac output power is found by substituting ICmax for IC1max to get

The total power supplied to the stage is the sum of the power to the transistor and the power to

the bias and compensation circuitry.

If we subtract the power to the load from the power supplied to the transistors, we find the

power being dissipated in the transistors the power dissipated by a single transistor is one

half of this value. Thus,

we are assuming that the base current is negligible.

The efficiency of the Class B push-pull amplifier is the ratio of the output power to the

power delivered to the transistor. Thus we neglect the power dissipated by the bias circuitry.

This amplifier is more efficient than a Class A amplifier. It is often used in output circuits

where efficiency important design requirement.

Therefore,

In choosing a transistor, it is important that the power rating is equal to or exceeds the

maximum power Pmax.



Class C amplifier:

A class C amplifier can produce more power than a class B amplifier.

Consider the case of a radio transmitter in which the audio signals are raised in their frequency

to the medium or short wave band to that they can be easily transmitted. The high frequency

introduced is in radio frequency range and it serves as the carrier of the audio signal. The

process of raising the audio signal to radio frequency called modulation.

The modulated wave has a relatively narrow band of frequencies centered around the carrier

frequencies. At any instant, there are several transmitter transmitting programmes

simultaneously. The radio receiver selects the signals of desired frequencies to which it is

tuned, amplifies it and converts it back to audio range. Therefore, tuned voltage amplifiers

are used. In short, the tuned voltage amplifiers selects the desired radio frequency signal out

of a number of RF signals present at that instant and then amplifies the selected RF signal to

the desired level as shown in fig. 1.

Fig. 1

Class C operation means that the collector current flows for less than 180° of the ac cycle. This

implies that the collector current of a class C amplifier is highly non-sinusoidal because

current flows in pulses. To avoid distortion, class C amplifier makes use of a resonant tank

circuit. This results in a sinusoidal output voltage.




The resonant tank circuit is tuned to the frequency of the input signal. When the circuit has a

high quality factor (Q) parallel resonance occurs at approximately

At the resonance frequency, the impedence of the parallel resonant circuit is very high as

shown in fig. 2 and is purely resistive. When the circuit is tuned to the resonant frequency,

the voltage across RL is maximum and sinusoidal. The higher the Q of the circuit, the faster

the gain drops off on either side of resonance

Fig. 2

Fig. 3

The dc equivalent circuit is shown fig. 3. No bias is applied to the transistor. Therefore, its Q-

point is at cut off on the dc load line VBE= 0.7V. Therefore no IL current flows until input is





more than 0.7 V. Also dc resistance is RS (the resistance of inductor) which is very small

and therefore dc load line is almost vertical. There is no danger of thermal runway because

there is no current other than from leakage.

The IC(sat) current is given by

where rC is the collector resistance.

The voltage VCE(sate) is given by

when ICQ = 0, VCEQ = VCC

When the Q of a resonant circuit is greater than 10. One can use the approximate ac equivalent

circuit. The series resistance of the inductor is lumped into the collector resistance. At

resonance, the peak-to-peak load voltage reaches a maximum. The bandwidth of a resonant

circuit is given by

Band width (BW) = f 2 f 1

f 1 = Lower cut off frequency.

f 2 = Upper cut off frequency

The bandwidth is related to the resonant frequency and the circuit Q as below:

BW = f r / Q

This means a large Q produces small BW equivalent to sharp tuning. These amplifiers have Q

greater than 10. This means that the BW is less than 10% of the resonant frequencies. These

amplifiers are also called narrow band amplifier.



When the tank circuit is resonant the ac load impedence seen by the collector current source is

purely resistive and the collector current is minimum. Above and below resonance, the ac

load impedence decreases and the collector current decreases. Any coil or inductor has some

series resistance RS as shown in fig. 4.

The Q of all coil is given by.

Q L = X L / R L

Fig. 4

The series resistance can be replaced by parallel resistance RP . This equivalent resistance is

given by

R P = Q L RL

Now all the losses in the coil are now being represented by the parallel resistance RP and series

resistance RS no longer exists XC cancels XL at resonance. Leaving only RP in parallel with

RL.

Therefore,

RC = RP || RL

Q of the overall circuit = rC / XL

At the resonance frequency, the impedence of the parallel resonant circuit is very high as

shown in fig. 2 and is purely resistive. When the circuit is tuned to the resonant frequency,

the voltage across RL is maximum and sinusoidal. The higher the Q of the circuit, the faster

the gain drops off on either side of resonance





Fig. 2

Fig. 3

The dc equivalent circuit is shown fig. 3. No bias is applied to the transistor. Therefore, its Q-

point is at cut off on the dc load line VBE= 0.7V. Therefore no IL current flows until input is

more than 0.7 V. Also dc resistance is RS (the resistance of inductor) which is very small

and therefore dc load line is almost vertical. There is no danger of thermal runway because

there is no current other than from leakage.

The IC(sat) current is given by

where rC is the collector resistance.




The voltage VCE(sate) is given by

when ICQ = 0, VCEQ = VCC

When the Q of a resonant circuit is greater than 10. One can use the approximate ac equivalent

circuit. The series resistance of the inductor is lumped into the collector resistance. At

resonance, the peak-to-peak load voltage reaches a maximum. The bandwidth of a resonant

circuit is given by

Band width (BW) = f 2 f 1

f 1 = Lower cut off frequency.

f 2 = Upper cut off frequency

The bandwidth is related to the resonant frequency and the circuit Q as below:

BW = f r / Q

This means a large Q produces small BW equivalent to sharp tuning. These amplifiers have Q

greater than 10. This means that the BW is less than 10% of the resonant frequencies. These

amplifiers are also called narrow band amplifier.

When the tank circuit is resonant the ac load impedence seen by the collector current source is

purely resistive and the collector current is minimum. Above and below resonance, the ac

load impedence decreases and the collector current decreases. Any coil or inductor has some

series resistance RS as shown in fig. 4.

The Q of all coil is given by.

Q L = X L / R L




Fig. 4

The series resistance can be replaced by parallel resistance RP . This equivalent resistance is

given by

R P = Q L RL

Now all the losses in the coil are now being represented by the parallel resistance RP and series

resistance RS no longer exists XC cancels XL at resonance. Leaving only RP in parallel with

RL.

Therefore,

RC = RP || RL

Q of the overall circuit = rC / XL

Current Sources

There are different methods of simulating a dc current source for integrated circuit amplifier

biasing. One type of current source used to provide a fixed current is the fixed bias transistor

circuit. The problem with this type of current source is that it requires too many resistors to

be practically implemented on IC. The resistors in the following circuits are small and easy

to fabricate on IC chips. When the current source is used to replace a large resistor the

Thevenin resistance of the current source is the equivalent resistance value.



Expected Questions

1. Explain the bandwidth of amplifier

2. Explain the ground reference of the amplifier

3. What are the h-parameters

4. Determine h-parameters of an amplifier

5. Explain CE amplifier with emitter resistance

UNIT-7

Oscillators



Basic operation of an Oscillator

An amplifier with positive feedback results in oscillations if the following

conditions are satisfied:

– The loop gain ( product of the gain of the amplifier and the gain of the

feedback network) is unity

– The total phase shift in the loop is 0

• If the output signal is sinusoidal, such a circuit is referred to as sinusoidal oscillator.

When the switch at the amplifier input is open, there are no oscillations. Imagine that a voltage

Vi is fed to the circuit and the switch is closed. This results in V

When the switch at the amplifier input is open, there are no oscillations. Imagine that a voltage

Vi is fed to the circuit and the switch is closed. This results in and

is fed back to the circuit. If we make Vf = V then even if we remove the input

voltage to the circuit, the output continues to exist.

then from the above equation, it is clear that, A =1V

Thus in the above block diagram, by closing the switch and removing the input, we are



able to get the oscillations at the output if AV =1, where A is called the Loop gain.

Positive feedback refers to the fact that the fed back signal is in phase with the input signal.

This means that the signal experiences 0 phase shift while traveling in the loop. The

above condition along with the unity loop gain needs to be satisfied to get the sustained

oscillations. These conditions are referred to as „Barkhausen criterion‟. Another way of

seeing how the feedback circuit provides operation as an oscillator is obtained by noting the

denominator in the basic equation

Af = A / (1+A).

When A = -1 or magnitude 1 at a phase angle of 180, the denominator becomes 0 and

the gain with feedback Af becomes infinite.Thus, an infinitesimal signal ( noise voltage) can

provide a measurable output voltage, and the circuit acts as an oscillator even without an

input signal.

Phase shift oscillator

• The phase

shift oscillator utilizes three RC circuits to provide 180º phase shift that when coupled with

the 180º of the op-amp itself provides the necessary feedback to sustain oscillations.

• The gain must be at least 29 to maintain the oscillations. The frequency of resonance for the

this type is similar to any RC circuit oscillator:



• The amplifier stage is self biased with a capacitor bypassed source resistor Rs and a drain

bias resistor Rd . The FET device parameters of interest are gm and rd.

• At the operating frequency, we can assume that the input impedance of the

amplifier is infinite .

• This is a valid approximation provided, the oscillator operating frequency is low enough so

that FET capacitive impedances can be neglected.

• The output impedance of the amplifier stage given by R should also be small compared to

the impedance seen looking into the feedback network so that no attenuation due to loading

occurs.

• If a transistor is used as the active element of the amplifier stage, the output of thfeedback

network is loaded appreciably by the relatively low input resistance ( h ie) of the transistor.



• An emitter – follower input stage followed by a common emitter amplifier stage

could be used.If a single transistor stage is desired, the use of voltage – shunt feedback is more

suitable. Here, the feedback signal is coupled through the feedback resistor R‟ in series with

the amplifier stage input resistance ( R i).

Problem:

It is desired to design a phase shift oscillator using an FET having g = 5000S rd= 40 k ,

and a feedback circuit value of R = 10 k. Select the value of C for oscillator operation at 5

kHz and RD for A > 29 to ensure oscillator action.

Wien bridge oscillator



A Wien bridge oscillator is a type of electronic oscillator that generates sine waves. It can

generate a large range of frequencies. The circuit is based on an electrical network originally

developed by Max Wien in 1891. The bridge comprises four resistors and two capacitors. It

can also be viewed as a positive feedback system combined with a bandpass filter. Wien did

not have a means of developing electronic gain so a workable oscillator could not be realized.

The modern circuit is derived from William Hewlett's 1939 Stanford University master's

degree thesis. Hewlett, along with David Packard co-founded Hewlett-Packard. Their first

product was the HP 200A, a precision sine wave oscillator based on the Wien bridge. The

200A was one of the first instruments to produce such low distortion.

The frequency of oscillation is given by:

Analysis

Input admittance analysis

If a voltage source is applied directly to the input of an ideal amplifier with feedback, the input

current will be:

http://en.wikipedia.org/wiki/Electronic_oscillator

http://en.wikipedia.org/wiki/Sine_wave

http://en.wikipedia.org/wiki/Frequencies

http://en.wikipedia.org/wiki/Electrical_network

http://en.wikipedia.org/wiki/Max_Wien

http://en.wikipedia.org/wiki/Bridge_circuit

http://en.wikipedia.org/wiki/Resistor

http://en.wikipedia.org/wiki/Capacitor

http://en.wikipedia.org/wiki/Bandpass_filter

http://en.wikipedia.org/wiki/Gain

http://en.wikipedia.org/wiki/William_Reddington_Hewlett

http://en.wikipedia.org/wiki/Stanford_University

http://en.wikipedia.org/wiki/David_Packard

http://en.wikipedia.org/wiki/Hewlett-Packard

http://en.wikipedia.org/wiki/Distortion



Where vin is the input voltage, vout is the output voltage, and Zf is the feedback impedance. If

the voltage gain of the amplifier is defined as:

And the input admittance is defined as:

Input admittance can be rewritten as:

For the Wien bridge, Zf is given by:

If Av is greater than 1, the input admittance is a negative resistance in parallel with an

inductance. The inductance is:

If a capacitor with the same value of C is placed in parallel with the input, the circuit has a

natural resonance at:

http://en.wikipedia.org/wiki/Admittance

http://en.wikipedia.org/wiki/Negative_resistance

http://en.wikipedia.org/wiki/Inductance

http://en.wikipedia.org/wiki/Resonance



Substituting and solving for inductance yields:

If Av is chosen to be 3:

Lin = R2C

Substituting this value yields:

Or:

Similarly, the input resistance at the frequency above is:

For Av = 3:

Rin = − R

If a resistor is placed in parallel with the amplifier input, it will cancel some of the negative

resistance. If the net resistance is negative, amplitude will grow until clipping occurs.

Similarly, if the net resistance is positive, oscillation amplitude will decay. If a resistance is

added in parallel with exactly the value of R, the net resistance will be infinite and the circuit

can sustain stable oscillation at any amplitude allowed by the amplifier.

Notice that increasing the gain makes the net resistance more negative, which increases

amplitude. If gain is reduced to exactly 3 when a suitable amplitude is reached, stable, low



distortion oscillations will result. Amplitude stabilization circuits typically increase gain until a

suitable output amplitude is reached. As long as R, C, and the amplifier are linear, distortion

will be minimal.

Crystal oscillator

A crystal oscillator is an electronic circuit that uses the mechanical resonance of a vibrating

crystal of piezoelectric material to create an electrical signal with a very precise frequency.

This frequency is commonly used to keep track of time (as in quartz wristwatches), to provide

a stable clock signal for digital integrated circuits, and to stabilize frequencies for radio

transmitters and receivers. The most common type of piezoelectric resonator used is the quartz

crystal, so oscillator circuits designed around them were called "crystal oscillators".

Quartz crystals are manufactured for frequencies from a few tens of kilohertz to tens of

megahertz. More than two billion (2×109) crystals are manufactured annually. Most are small

devices for consumer devices such as wristwatches, clocks, radios, computers, and cellphones.

Quartz crystals are also found inside test and measurement equipment, such as counters, signal

generators, and oscilloscopes

A quartz crystal can be modeled as an electrical network with a low impedance (series) and a

high impedance (parallel) resonance point spaced closely together. Mathematically (using the

Laplace transform) the impedance of this network can be written as:

or,

where s is the complex frequency (s = jω), ωs is the series resonant frequency in radians per

second and ωp is the parallel resonant frequency in radians per second.

Adding additional capacitance across a crystal will cause the parallel resonance to shift

downward. This can be used to adjust the frequency at which a crystal oscillator oscillates.

Crystal manufacturers normally cut and trim their crystals to have a specified resonance

http://en.wikipedia.org/wiki/Electronic_circuit

http://en.wikipedia.org/wiki/Resonance

http://en.wikipedia.org/wiki/Crystal

http://en.wikipedia.org/wiki/Piezoelectricity#Materials

http://en.wikipedia.org/wiki/Frequency

http://en.wikipedia.org/wiki/Quartz_clock

http://en.wikipedia.org/wiki/Clock_signal

http://en.wikipedia.org/wiki/Digital

http://en.wikipedia.org/wiki/Integrated_circuits

http://en.wikipedia.org/wiki/Radio_transmitter



http://en.wikipedia.org/wiki/Radio_receiver

http://en.wikipedia.org/wiki/Quartz_crystal



http://en.wikipedia.org/wiki/Kilohertz

http://en.wikipedia.org/wiki/Wristwatch

http://en.wikipedia.org/wiki/Clock

http://en.wikipedia.org/wiki/Radio

http://en.wikipedia.org/wiki/Computer

http://en.wikipedia.org/wiki/Cellphone

http://en.wikipedia.org/wiki/Signal_generator



http://en.wikipedia.org/wiki/Oscilloscope

http://en.wikipedia.org/wiki/Electrical_impedance

http://en.wikipedia.org/wiki/Electrical_impedance

http://en.wikipedia.org/wiki/Laplace_transform

http://en.wikipedia.org/wiki/Radian

http://en.wikipedia.org/wiki/Capacitance



frequency with a known 'load' capacitance added to the crystal. For example, a 6 pF 32 kHz

crystal has a parallel resonance frequency of 32,768 Hz when a 6.0 pF capacitor is placed

across the crystal. Without this capacitance, the resonance frequency is higher than 32,768 Hz.

Expected Questions

1. Explain basic operation of phase shift oscillator

2. Design RC phase shift oscillator using BJT

3. It is desired to design a phase shift oscillator using an FET having g = 5000S

rd= 40 k , and a feedback circuit value of R = 10 k. Select the value of C for

oscillator operation at 5 kHz and RD for A > 29 to ensure oscillator action.

4. Obtain expression for closed loop gain of non-inverting amplifier

5. With necessary sketch and characteristics curves explain the operation of a

Schmitt trigger

6. With a neat circuit diagram , explain the working principle of RC phase shift

oscillator, with relevant equations

7. What are the tuned oscillator? Explain one type of oscillator

8. Explain how feedback can be used as oscillator

9. Explain with the help of circuit diagram, the working of an RC phase shift oscillator

Chapter-8

Field Effect Transistor

The field effect transistor is a semiconductor device, which depends for its operation on the

control of current by an electric field. There are two of field effect transistors:

1. JFET (Junction Field Effect Transistor)

2. MOSFET (Metal Oxide Semiconductor Field Effect Transistor)

http://en.wikipedia.org/wiki/Capacitor



The FET has several advantages over conventional transistor.

1. In a conventional transistor, the operation depends upon the flow of majority and

minority carriers. That is why it is called bipolar transistor. In FET the operation depends

upon the flow of majority carriers only. It is called unipolar device.

2. The input to conventional transistor amplifier involves a forward biased PN junction

with its inherently low dynamic impedance. The input to FET involves a reverse biased PN

junction hence the high input impedance of the order of M-ohm.

3. It is less noisy than a bipolar transistor.

4. It exhibits no offset voltage at zero drain current.

5. It has thermal stability.

6. It is relatively immune to radiation.

The main disadvantage is its relatively small gain bandwidth product in comparison with

conventional transistor.

Operation of FET:

Consider a sample bar of N-type semiconductor. This is called N-channel and it is electrically

equivalent to a resistance as shown in fig. 1

Ohmic contacts are then added on each side of the channel to bring the external connection.

Thus if a voltage is applied across the bar, the current flows through the channel.

The terminal from where the majority carriers (electrons) enter the channel is called source

designated by S. The terminal through which majority carriers leaves the channel is called

drain and designated by D. For an N-channel device, electrons are the majority carriers.

Hence the circuit behaves like a dc voltage VDS applied across a resistance RDS. The

resulting current is the drain current ID. If VDS increases, ID increases proportionally.

Now on both sides of the n-type bar heavily doped regions of p-type impurity have been

formed by any method for creating pn junction. These impurity regions are called gates

(gate1 and gate2) as shown in fig. 2.





Both the gates are internally connected and they are grounded yielding zero gate source voltage

(VGS =0). The word gate is used because the potential applied between gate and source

controls the channel width and hence the current.

As with all PN junctions, a depletion region is formed on the two sides of the reverse biased

PN junction. The current carriers have diffused across the junction, leaving only uncovered

positive ions on the n side and negative ions on the p side. The depletion region width

increases with the magnitude of reverse bias. The conductivity of this channel is normally

zero because of the unavailability of current carriers.

The potential at any point along the channel depends on the distance of that point from the

drain, points close to the drain are at a higher positive potential, relative to ground, then

points close to the source. Both depletion regions are therefore subject to greater reverse

voltage near the drain. Therefore the depletion region width increases as we move towards

drain. The flow of electrons from source to drain is now restricted to the narrow channel

between the no conducting depletion regions. The width of this channel determines the

resistance between drain and source.

Fig. 2

Consider now the behavior of drain current ID vs drain source voltage VDS. The gate source

voltage is zero therefore VGS= 0. Suppose that VDS is gradually linearly increased linearly

from 0V. ID also increases.

Since the channel behaves as a semiconductor resistance, therefore it follows ohm's law. The

region is called ohmic region, with increasing current, the ohmic voltage drop between the

source and the channel region reverse biased the junction, the conducting portion of the



channel begins to constrict and ID begins to level off until a specific value of VDS is reached,

called the pinch of voltage VP.

At this point further increase in VDS do not produce corresponding increase in ID. Instead, as

VDS increases, both depletion regions extend further into the channel, resulting in a no more

cross section, and hence a higher channel resistance. Thus even though, there is more

voltage, the resistance is also greater and the current remains relatively constant. This is

called pinch off or saturation region. The current in this region is maximum current that FET

can produce and designated by IDSS. (Drain to source current with gate shorted).

Fig. 3

As with all pn junctions, when the reverse voltage exceeds a certain level, avalanche

breakdown of pn junction occurs and ID rises very rapidly as shown in fig. 3.

Consider now an N-channel JFET with a reverse gate source voltage as shown in fig. 4.

Fig. 4

Fig. 5





The additional reverse bias, pinch off will occur for smaller values of | VDS |, and the maximum

drain current will be smaller. A family of curves for different values of VGS(negative) is

shown in fig. 5.

Suppose that VGS= 0 and that due of VDS at a specific point along the channel is +5V with

respect to ground. Therefore reverse voltage across either p-n junction is now 5V. If VGS is

decreased from 0 to ?1V the net reverse bias near the point is 5 - (-1) = 6V. Thus for any

fixed value of VDS, the channel width decreases as VGS is made more negative.

Thus ID value changes correspondingly. When the gate voltage is negative enough, the

depletion layers touch each other and the conducting channel pinches off (disappears). In

this case the drain current is cut off. The gate voltage that produces cut off is symbolized

VGS(off) . It is same as pinch off voltage.

Since the gate source junction is a reverse biased silicon diode, only a very small reverse

current flows through it. Ideally gate current is zero. As a result, all the free electrons from

the source go to the drain i.e. ID = IS. Because the gate draws almost negligible reverse

current the input resistance is very high 10's or 100's of M ohm. Therefore where high input

impedance is required, JFET is preferred over BJT. The disadvantage is less control over

output current i.e. FET takes larger changes in input voltage to produce changes in output

current. For this reason, JFET has less voltage gain than a bipolar amplifier.

Biasing the Field Effect Transistor

Transductance Curves:

The transductance curve of a JFET is a graph of output current (ID) vs input voltage (VGS) as

shown in fig. 1.





Fig. 1

By reading the value of ID and VGS for a particular value of VDS, the transductance curve can

be plotted. The transductance curve is a part of parabola. It has an equation of

Data sheet provides only IDSS and VGS(off) value. Using these values the transductance curve

can be plotted.

Biasing the FET:

The FET can be biased as an amplifier. Consider the common source drain characteristic of a

JFET. For linear amplification, Q point must be selected somewhere in the saturation region.

Q point is selected on the basis of ac performance i.e. gain, frequency response, noise,

power, current and voltage ratings.

Gate Bias:

Fig. 2, shows a simple gate bias circuit.

Fig. 2

Separate VGS supply is used to set up Q point. This is the worst way to select Q point. The

reason is that there is considerable variation between the maximum and minimum values of

FET parameters e.g.




IDSS VGS(off)

Minimum 4mA -2V

Maximum 13mA -8V

This implies that the minimum and maximum transductance curves are displaced as shown in

fig. 3.

Gate bias applies a fixed voltage to the gate. This fixed voltage results in a Q point that is

highly sensitive to the particular JFET used. For instance, if VGS= -1V the Q point may very

from Q1 to Q2 depending upon the JFET parameter is use.

At Q1, ID= 0.016 (1 - (1/8))2 = 12.3 mA

At Q2, ID= 0.004 (1-(1/2))2 = 1 mA.

The variation in drain current is very large.

Self Bias: Fig. 4, shows a self bias circuit another way to bias a FET. Only a drain supply is

used and no gate supply. The idea is to use the voltage across RS to produce the gate source

reverse voltage.

This is a form of a local feedback similar to that used with bipolar transistors. If drain current

increases, the voltage drop across RS increases because the ID RS increases. This increases

the gate source reverse voltage which makes the channel narrow and reduces the drain

current. The overall effect is to partially offset the original increase in drain current.

Similarly, if ID decreases, drop across RS decreases, hence reverse bias decreases and ID

increases.





Fig. 4

Since the gate source junction is reverse biased, negligible gate current flows through RG and

so the gate voltage with respect to ground is zero.

VG= 0;

The source to ground voltage equals the product of the drain current and the source resistance.

VS= ID R S.

The gate source voltage is the difference between the gate voltage and the source voltage.

VGS = VG ? VS = 0 ? IDRS

VGS = -ID RS.

This means that the gate source voltage equals the negative of the voltage across the source

resistor. The greater the drain current, the more negative the gate source voltage becomes.

Rearranging the equation:

ID = -VGS / RS

The graph of this equation is called self base line a shown in Fig. 5.




Fig. 5

The operating point on transductance curve is the intersection of self bias line and

transductance curve. The slope of the line is (-1 / RS). If the source resistance is very large (-

1 / RS is small) then Q-point is far down the transductance curve and the drain current is

small. When RS is small, the Q point is far up the transductance curve and the drain current

is large. In between there is an optimum value of RS that sets up a Q point near the middle of

the transductance curve.

The transductance curve varies widely for FET (because of variation in IDSS and VGS(off)) as

shown in fig. 6. The actual curve may be in between there extremes. A and B are the

optimum points for the two extreme curves. To find the optimum resistance RS, so that Q-

point is correct for all the curves, A and B points are joined such that it passes through

origin.

The slope of this line gives the resistance value RS( VGS = -ID RS). The current IQ is such that IA

> IQ > IB. Here A, Q and B all points are in straight line.




Fig. 6

Consider the case where a line drawn to pass between points A and B does not pass through the

origin. The equation VGS = - ID RS is not valid. The equation of this line is VGS = VGG ? ID

RS.

Such a bias relationship may be obtained by adding a fixed bias to the gate in addition to the

source self bias as shown in fig. 7.

Fig. 7

In this circuit.

VGG = RS IG + VGS + ID RS




Since RS IG = 0;

VGG = VGS + ID RS

or VGS = VGG? ID RS

Voltage Divider Bias :

The biasing circuit based on single power supply is shown in fig. 1. This is similar to the

voltage divider bias used with a bipolar transistor.

Fig. 1

The Thevenin voltage VTH applied to the gate is

The Thevenin resistance is given as

The gate current is assumed to be negligible. VTH is the dc voltage from gate to ground.




The drain current ID is given by

and the dc voltage from the drain to ground is VD = VDD ? ID RD.

If VTH is large enough to swamp out VGS the drain current is approximately constant for any

JFET as shown in fig. 2.

Fig. 2

There is a problem in JFET. In a BJT, VBE is approximately 0.7V, with only minor variations

from one transistor to other. In a FET, VGS can vary several volts from one JFET to another.

It is therefore, difficult to make VTH large enough to swamp out VGS. For this reason,

voltage divider bias is less effective with, FET than BJT. Therefore, VGS is not negligible.

The current increases slightly from Q2 to Q1. However, voltage divider bias maintains ID

nearly constant.

Consider a voltage divider bias circuit shown in fig. 3.





Difference in ID (min) and ID (max) is less

VD (max) = 30 ? 2.13 * 4.7 = 20 V

VD (min) = 30 ? 2.67 * 4.7 = 17.5 V

Fig. 3

Current Source Bias:

This is another way to produce solid Q point. The aim is to produce a drain current that is

independent of VGS. Voltage divider bias and self bias attempt to do this by swamping out of

variations in VGS.

Using two power supplies:

The current source bias can be used to make ID constant fig. 4.

Fig. 4




The bipolar transistor is emitter biased; its collector current is given by

IC = (VEE ? VBE ) / RE.

Because the bipolar transistor acts like a current source, it forces the drain current to equal the

bipolar collector current.

ID = IC

Since IC is constant, both Q points have the same value of drain current. The current source

effectively wipes out the influence of VGS. Although VGS is different for each Q point, it no

longer influences the value of drain current.

Using One power supply:

When only a positive supply is available, the circuit shown in fig. 5, can be used to set up a

constant drain current.

In this case, the bipolar transistor is voltage divider biased. Assuming a stiff voltage divider,

the emitter and collector currents are constant for all bipolar transistors. This forces the FET

drain current equal the bipolar collector current.

Fig. 5

Transductance: The transductance of a FET is defined as




Because the changes in ID and VGS are equivalent to ac current and voltage. This equation can

be written as

The unit of gm is mho or siemems.

Typical value of gm is 2000 m A / V.

Fig. 6

The value of gm can be obtained from the transductance curve as shown in fig. 6.

If A and B points are considered, than a change in VGS produces a change in ID. The ratio of ID

and VGS is the value of gm between A and B points. If C and D points are considered, then

same change in VGS produces more change in ID. Therefore, gm value is higher. In a nutshell,

gm tells us how much control gate voltage has over drain current. Higher the value of gm, the

more effective is gate voltage in controlling gate current. The second parameter rd is the

drain resistance.




FET a amplifier

Similar to Bipolar junction transistor. JFET can also be used as an amplifier. The ac equivalent

circuit of a JFET is shown in fig. 1.

Fig. 1

The resistance between the gate and the source RGS is very high. The drain of a JFET acts like

a current source with a value of gm Vgs. This model is applicable at low frequencies.

From the ac equivalent model

The amplification factor µ for FET is defined as

When VGS = 0, gm has its maximum value. The maximum value is designated as gmo. Again

consider the equation,




As VGS increases, gm decreases linearly.

Measuring IDSS and gm, VGS(off) can be determined

FET as Amplifier:

Fig. 2, shows a common source amplifier.

Fig. 2

When a small ac signal is coupled into the gate it produces variations in gate source voltage.

This produces a sinusoidal drain current. Since an ac current flows through the drain




resistor. An amplified ac voltage is obtained at the output. An increase in gate source

voltage produces more drain current, which means that the drain voltage is decreasing.

Since the positive half cycle of input voltage produces the negative half cycle of output

voltage, we get phase inversion in a CS amplifier.

The ac equivalent circuit is shown in fig. 3.

Fig. 3

The ac output voltage is

vout = - gm v gS RD

Negative sign means phase inversion. Because the ac source is directly connected between the

gate source terminals therefore ac input voltage equals

Vin = Vgs

The voltage gain is given by

The further simplified model of the amplifieris shown in fig. 4.





Fig. 4

Zin is the input impedance. At low frequencies, this is parallel combination of R1|| R2|| RGS.

Since RGS is very large, it is parallel combination of R1 & R2. A Vin is output voltage and RD

is the output impedance.

Because of nonlinear transductance curve, a JFET distorts large signals, as shown in fig. 5.

Given a sinusoidal input voltage, we get a non-sinusoidal output current in which positive half

cycle is elongated and negative cycle is compressed. This type of distortion is called Square

law distortion because the transductance curve is parabolic.

This distortion is undesirable for an amplifier. One way to minimize this is to keep the signal

small. In that case a part of the curve is used and operation is approximately linear. Some times

swamping resistor is used to minimize distortion and gain constant. Now the source is no

longer ac ground as shown in fig. 6.





The drain current through rS produces an ac voltage between the source and ground. If rS is

large enough the local feedback can swamp out the non-linearity of the curve. Then the

voltage gain approaches an ideal value of RD / rS.

Since RGS approaches infinity therefore, all the drain current flows through rS producing a

voltage drop of gm VgS rS. The ac equivalent circuit is shown in fig. 7.

Fig. 7

The voltage gain reduces but voltage gain is less effective by change in gm. rS must be greater

than 1 / gm only then




Expected questions

1. What are the advantages of FET

2. Explain the operation of FET

3. Draw transconductance curve of FET

4. Determine gm for an n-channel JFET with characteristic curve shown in fig. 1.

Fig. 1

5. Determine g m for a JFET where IDSS = 7 mA, VP = -3.5 V and VDD = 15V. Choose

a reasonable location for the Q-point.

6. Discuss the difference between FET and BJT

7. Explain FET small signal model with help of graphical representation of gm

8. With necessary oc equivalent model of JFET common-drain configuration. Obtain

the expression for Zi,Zo, Av

9. Differentiate between enhancement and depletion MOSFET


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